Chapter 2, Problem 128P

The viscosity of some fluids changes when a strong electric ?eld is applied on them. This phenomenon is known as the electrorheolegical (ER) effect, and fluids that exhibit such behavior are known as ER fluids. The Bingham plastic model for shear stress, which is expressed as $\tau ={\tau }_y+\mu \left(du/dy\right)$ is widely used to describe ER fluid behavior because of its simplicity. One of the most promising applications of ER fluids is the ER Clutch. A typical multidisk ER Clutch consists of several equally spaced steel disks of inner radius $R_1$ and outer radius $R_2,$N of them attached to the input shaft. The gap h between the parallel disks is filled with a viscous fluid. (a) Find a relationship for the torque generated by the clutch when the output shaft is stationary and (b) calculate the torque for an ER clutch with $N=11$ for $R_1=50\text{mm,}{R}_2=200\text{mm,}$ $\eta =2400\text{rpm}$ if the fluids is SAE 10 with $\mu =0.1\text{Pa}\cdot \text{s,}{\tau }_y=2.5\text{kPa,}$ and $h=1.2\text{mm}\text{.}$

Answer: (b) $2060\text{N}\cdot \text{m}$

To determine

(a)

The relationship for the torque generated by a clutch when the output shaft is stationary.

Answer to Problem 128P

The torque generated by a clutch when the output shaft is stationary is $4\pi N\left[\frac{{\tau }_y}{3}\left(R_2^3-R_1^3\right)+\frac{\mu \omega }{4h}\left(R_2^4-R_1^4\right)\right]$.

Explanation of Solution

Given information:

The thickness of the fluid is $h$, the inner radius is $R_1$, the outer radius is $R_2,$ and the number of plate in clutch is $N$.

Write the expression for the shear stress for Bingham plastic.

  $\tau ={\tau }_y+\mu \left(\frac{du}{dy}\right)$...... (I)

Here, the shear stress is $\tau$, the shear stress in $y$ direction is ${\tau }_y$, the dynamic viscosity is $\mu$ and the speed of the fluid is $u$.

Write the expression for the velocity gradient.

  $\frac{V}{h}=\frac{du}{dy}$...... (II)

Here, the thickness of the film is $h$ and the velocity of shaft is $V$.

Write the expression for the velocity of shaft.

  $V=r\omega$...... (III)

Here, the angular velocity is $\omega$ and the radius of the shaft is $r$.

Write the expression for the force.

  $dF=\tau dA$...... (IV)

Here, the force is $dF$ and the area of the plate is $dA$.

Write the expression for the torque generated by the shaft.

  $dT=rdF$...... (V)

Here, the torque generated by the shaft is $dT$.

Write the expression for the area of the plate.

  $dA=2\pi rdr$...... (VI)

Calculation:

Substitute $\frac{V}{h}$ for $\frac{du}{dy}$ in Equation (I).

  $\tau ={\tau }_y+\mu \left(\frac{V}{h}\right)$...... (VII)

Substitute $r\omega$ for $V$ in Equation (VII).

  $\tau ={\tau }_y+\mu \left(\frac{r\omega }{h}\right)$

Substitute ${\tau }_y+\mu \left(\frac{r\omega }{h}\right)$ for $\tau$ and $2\pi rdr$ for $dA$ in Equation (IV).

  $dF=\left[{\tau }_y+\mu \left(\frac{r\omega }{h}\right)\right]\left(2\pi rdr\right)$

Substitute $\left[{\tau }_y+\mu \left(\frac{r\omega }{h}\right)\right]\left(2\pi rdr\right)$ for $dF$ in Equation (V).

  $dT=r\left[{\tau }_y+\mu \left(\frac{r\omega }{h}\right)\right]\left(2\pi rdr\right)$

Integerating both side taking lower limit $R_1$ and higher limit $R_2$ for $dr$.

  $\begin{array}{l}{\displaystyle \int dT}={\displaystyle \underset{R_1}{\overset{R_2}{\int }}r\left[{\tau }_y+\mu \left(\frac{r\omega }{h}\right)\right]\left(2\pi rdr\right)}\\ T=2\pi \left({\tau }_y{\displaystyle \underset{R_1}{\overset{R_2}{\int }}{r}^{2}dr}+\frac{\mu \omega }{h}{\displaystyle \underset{R_1}{\overset{R_2}{\int }}{r}^{3}dr}\right)\\ T=2\pi \left[\frac{{\tau }_y}{3}\left(R_2^3-R_1^3\right)+\frac{\mu \omega }{4h}\left(R_2^4-R_1^4\right)\right]\end{array}$

The torque generated by the both surface of the plate.

  $\begin{array}{c}T=2\left\{2\pi \left[\frac{{\tau }_y}{3}\left(R_2^3-R_1^3\right)+\frac{\mu \omega }{4h}\left(R_2^4-R_1^4\right)\right]\right\}\\ =4\pi \left[\frac{{\tau }_y}{3}\left(R_2^3-R_1^3\right)+\frac{\mu \omega }{4h}\left(R_2^4-R_1^4\right)\right]\end{array}$

For $N$ number of clutch plates the torque generated is

  $T=4\pi N\left[\frac{{\tau }_y}{3}\left(R_2^3-R_1^3\right)+\frac{\mu \omega }{4h}\left(R_2^4-R_1^4\right)\right]$...... (VIII)

Conclusion:

The torque generated by a clutch when the output shaft is stationary is $4\pi N\left[\frac{{\tau }_y}{3}\left(R_2^3-R_1^3\right)+\frac{\mu \omega }{4h}\left(R_2^4-R_1^4\right)\right]$.

To determine

(b)

The torque for an ER clutch.

Answer to Problem 128P

The torque for an ER clutch is $2060.54\text{N}\cdot \text{m}$.

Explanation of Solution

Given information:

The number of clutch plate is $11$, the inner radius of plate is $50\text{mm}$, the outer radius is $200\text{mm}$, the speed of input shaft is $2400\text{rpm}$, the viscosity is $0.1\text{Pa}\cdot \text{s}$, the fluid thickness is $1.2\text{mm,}$ and the stress on the plate is $2.5\text{kPa}$.

Write the expression for the angular speed of the shaft.

  $\omega =\frac{2\pi \dot{n}}{60}$..... (IX)

Here, the angular speed is $\omega$ and the speed of the shaft is $\dot{n}$.

Calculation:

Substitute $2400\text{rpm}$ for $\dot{n}$ in Equation (IX).

  $\begin{array}{c}\omega =\frac{2\pi \times \left(2400\text{rpm}\right)}{60}\\ =\frac{15079.64}{60}\text{rad}/\text{s}\\ =251.32\text{rad}/\text{s}\end{array}$

Substitute $11$ for $N$, $2.5\text{kPa}$ for ${\tau }_y$, $200\text{mm}$ for $R_2$, $50\text{mm}$ for $R_1$, $0.1\text{Pa}\cdot \text{s}$ for $\mu$, $251.32\text{rad}/\text{s}$ for $\omega$ and $1.2\text{mm}$ for $h$ in Equation (VIII).

  $\begin{array}{c}T=4\pi \times 11\left[\begin{array}{l}\frac{2.5\text{kPa}}{3}\left({\left(200\text{mm}\right)}^3-{\left(50\text{mm}\right)}^3\right)\\ +\frac{\left(0.1\text{Pa}\cdot \text{s}\right)\left(251.32\text{rad}/\text{s}\right)}{4\times 1.2\text{mm}}\left({\left(200\text{mm}\right)}^4-{\left(50\text{mm}\right)}^4\right)\end{array}\right]\\ =4\pi \times 11\left[\begin{array}{l}\left(\left(0.83333\text{kPa}\right)\times \left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\right)\left(\left(7875000{\text{mm}}^3\right)\times \left(\frac{1{\text{m}}^3}{{10}^9{\text{mm}}^3}\right)\right)\\ +\frac{25.132\text{Pa}}{\left(\left(4.8\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}\left(\left(1593750000{\text{mm}}^4\right)\times \left(\frac{1{\text{m}}^4}{{10}^{12}{\text{mm}}^4}\right)\right)\end{array}\right]\\ =4\pi \times 11\left(14.9066\text{Pa}\cdot {\text{m}}^3\times \frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)\\ =2060.54\text{N}\cdot \text{m}\end{array}$

Conclusion:

The torque for an ER clutch is $2060.54\text{N}\cdot \text{m}$.